51问答网 > △ABC中，已知（a*a+b*b）sin(A+B)=(a*a–b*b)sin(A+B),判断△形状

△ABC中，已知（a*a+b*b）sin(A+B)=(a*a–b*b)sin(A+B),判断△形状

2025-04-05 18:25:29
推荐回答（2个）
回答1：

用到的公式:
sin（α+β）=sinαcosβ+cosαsinβ；
sin（α-β）=sinαcosβ-cosαsinβ；
正弦定理:
sinA/a=sinB/b=sinC/c=k
二倍角公式:
sin2α=2sinαcosα
解:
由（a²+b²）sin(A-B)=(a²-b²）sin(A+B)，
按公式展开得
（a²+b²)[sinAcosB-cosAsinB]=(a²-b²)[sinAcosB+cosAsinB]
移项（a²+b²)sinAcosB-(a²-b²)sinAcosB=（a²+b²)cosAsinB+(a²-b²)cosAsinB
合并得：2b²sinAcosB=2a²cosAsinB
由正弦定理得：sin²BsinAcosB=sin²AcosAsinB
sinBcosB=sinAcosA,
sin2B=sin2A,
得：2A=2B，或者2A=π-2B
即A=B或A+B=π/2
所以△ABC是等腰或直角

回答2：

题目错了吧

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